Exponential/logarithmic equation

[Chuckster]

Homework Statement

How many numbers, where x is is a whole number, satisfy the equation.

Homework Equations

$$x*3^{log_{x}5}=15$$

The Attempt at a Solution

Most of my attempts have been blocked due to the fact that i don't know what to do with the x that is not in the base of algorithm. I tried solving that, but failed... So, I'm asking your help guys!

 

[micromass]

This isn't really a calculus & Beyond question, but hey...

You could try to change the base of the logarithm. Do you have seen formula's that allow you to change the base of the logarithm??

 

[Chuckster]

This isn't really a calculus & Beyond question, but hey...

You could try to change the base of the logarithm. Do you have seen formula's that allow you to change the base of the logarithm??

I know those formulas, but i didn't find a way to properly use them in this case.

Also, I'm sorry if i posted this inf the wrong forum. Mods can move the topic.

 

[micromass]

What are the formula's then?? And why do you think they don't apply in this case?

 

[HallsofIvy]

$log_x(5)$ is only defined for x a positive number. If you are given that x is an integer, then the best way to solve this is "trial and error". Since 15 is not a very large number, x must be a relatively small integer. Just try x= 2, 3, 4, etc. until you find a solution.

Use, as micromass suggested, the formula
$$log_x(5)= \frac{log(5)}{log(x)}$$

Of course, I don't guarantee that the solution is an integer- you were the one who said that. Ah, wait the problem was not to find integer solutions but to tell how many there are!

 

[micromass]

Use, as micromass suggested, the formula
$$log_x(5)= \frac{log(5)}{log(x)}$$

It doesn't really matter that much, but if you take your logarithms in base 3, then it becomes even easier...

 

[Chuckster]

What are the formula's then?? And why do you think they don't apply in this case?

I could write $$3^{log_{x}5}$$ as $$3^\frac{1}{log_{5}x}$$ or $$3^\frac{log_{c}5}{log_{c}x}$$, where c is some other constant, but i don't know what to do with the x that multiplies 3, that's what's causing me trouble.

 

[micromass]

I could write $$3^{log_{x}5}$$ as $$3^\frac{1}{log_{5}x}$$ or $$3^\frac{log_{c}5}{log_{c}x}$$, where c is some other constant, but i don't know what to do with the x that multiplies 3, that's what's causing me trouble.

That last formula is the one to use. The x that multiplies the 3 isn't really the problem, it's the x inside the log inside the exponent that's the problem.

Now, to go on, do you know what

$$3^{\log_3(x)}$$

is?? It becomes a very easy expression. Use this!

 

[Chuckster]

That last formula is the one to use. The x that multiplies the 3 isn't really the problem, it's the x inside the log inside the exponent that's the problem.

Now, to go on, do you know what

$$3^{\log_3(x)}$$

is?? It becomes a very easy expression. Use this!

$$3^\frac{log_{3}5}{log_{3}x}$$ would be $$\frac{log_{3}5}{x}$$ because $$3^{\log_3(x)}$$ is $$x$$?

 

[HallsofIvy]

I recommend you go by "10"s to start with- evaluate $xe^{log_3(5)}$ for x= 10, 20, 30, 40, etc., comparing the values to 15, until you see something "happen". Then shift to "1"s.

 

[micromass]

Hmm, I see the problem...
I actually don't find any trivial ways to solve this problem.

What you perhaps can do is to find the first derivative of the function and calculate it's local extrema and where it is increasing/decreasing.

This should get you enough information to see how many solutions there are to your problem. With a little guessing, you can even find the solutions.

Or you can also use Halls approach...

Intriguing problem. Maybe this becomes a calculus question after all

 

[Chuckster]

Hmm, I see the problem...
I actually don't find any trivial ways to solve this problem.

What you perhaps can do is to find the first derivative of the function and calculate it's local extrema and where it is increasing/decreasing.

This should get you enough information to see how many solutions there are to your problem. With a little guessing, you can even find the solutions.

Or you can also use Halls approach...

Intriguing problem. Maybe this becomes a calculus question after all

I appreciate you all helping.
I see some of you suggested hit/miss option, that i want to avoid.

I'm interested if anyone could provide a procedure that's not based on guessing (but if we're at it - 5 is one of the solutions :D).

I'll keep on trying and if i come up with something, I'm letting you know. And i expect the same from you :D

 

[micromass]

I appreciate you all helping.
I see some of you suggested hit/miss option, that i want to avoid.

I'm interested if anyone could provide a procedure that's not based on guessing (but if we're at it - 5 is one of the solutions :D).

I'll keep on trying and if i come up with something, I'm letting you know. And i expect the same from you :D

Well, the question you posted only asks about the number of solutions, it didn't ask what the solutions were.

If you're only interested in the number of solutions, then I guess that my approach works fine. Find the first derivative, calculate the local extrema and where the function increases/decreases. Then you can find explicitly how many solutions there are! (there are more then 1).

Finding the explicit solutions, I think can only be done by guessing... This reminds me of following question:

Find the solutions to $$x=2^x$$. This kind of problems can only be done by guessing, you won't be able to solve this explicitly. (of course, we can use Lambert's W-function, but I don't count that as explicitly finding the solution).

So, I'll be very surprised if there is a general method to find the solutions of such an equation. However, there is a general method to find the number of solutions!

 

[Chuckster]

After a while, i finaly got to this equation, and solved it.
It was quite easy.

$$x*3^{log_{x}5}=15$$ /multiply this with log of base 3
$$log_{3}x + log_{x}5 = log_{3}5 + 1$$
when you convert
$$log_{x}5$$
to base 3, and and multiply the whole equation with
$$log_{3}x$$
and sort it out, factorize, you get:
$$(log_{3}x -1)(log_{x}5 - log_{3}5)=0$$
and therefore, x can be 3 or 5